II gimnazjum

Matematyka z plusem 2

a) (x+3)2=(x+3)(x+3)=x⋅x+x⋅3+3⋅x+3⋅3=x2+6x+9 b) (a−2)2=(a−2)(a−2)=a⋅a−2a−2a+4=a2−4a+4 c) (2x+5)2=(2x+5)(2x+5)=2x⋅2x+2x⋅5+5⋅2x+5⋅5=4x2+10x+10x+25=4x2+20x+25

Rozwiązanie 1

P□​=a2

Pole trójkąta prostokątnego to połowa iloczynu jego przyprostokątnych.

pierwsza liczba: 10-x druga liczba: 20-(10-x)=20-10+x=10+x (10−x)2+(10+x)2=100−10x−10x+x2+100+10x+10x+x2=200+2x2

a) (x−1)(x3+x2+x+1)=x⋅x3+x⋅x2+x⋅x+x⋅1−x3−x2−x−1=x4+x3+x2+x−x3−x2−x−1=x4−1

a) (x+1)(x+2)(x+3)=(x2+2x+x+2)(x+3)=(x2+3x+2)(x+3)=x2⋅x+x2⋅3+3x⋅x+3x⋅3+2x+2⋅3=x3+6x2+11x+6

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